Question

Integrated Concepts A basketball player jumps straight up for a ball. To do this, they lower their body 0.300 m and then accelerate through this distance by forcefully straightening their legs. This player leaves the floor with a vertical velocity sufficient to carry the player 0.900 m above the floor. (a) Calculate the player's velocity when they leave the floor. (b) Calculate their acceleration while they are straightening their legs. They go from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force they exert on the floor to do this, given that their mass is 110 kg.Calculate the player's velocity when leaving the floor in a vertical jump to reach a height of 0.900 m.

a) 4.43 m/s
b) 5.67 m/s
c) 6.81 m/s
d) 8.12 m/s

Answer 1

To find the player's velocity when leaving the floor, we can use the principle of conservation of energy. The acceleration while straightening their legs can be determined using the equation for final velocity. The force exerted on the floor can be calculated using Newton's second law of motion. The correct option is c) 6.81 m/s.

Step-by-step explanation:

To calculate the player's velocity when they leave the floor, we can use the principle of conservation of energy. The player initially possesses gravitational potential energy since they are 0.300 m above the floor, and this is converted to kinetic energy as they leave the floor. Using the equation:

Initial gravitational potential energy = final kinetic energy

mgh = (1/2)mv^2

where m is the mass of the player, g is the acceleration due to gravity, h is the initial height, and v is the velocity when leaving the floor.

By substituting the given values, we can solve for v:

v = sqrt(2gh)

where g is approximately 9.8 m/s^2. Substituting the given values, we find v ≈ 4.43 m/s.

To calculate the acceleration while the player is straightening their legs, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity (which is 0 m/s), a is the acceleration, and s is the distance traveled.

Substituting the given values, we can solve for a:

a = (v^2 - u^2) / (2s)

By substituting the given values, we find a ≈ 5.67 m/s^2.

Finally, to calculate the force exerted on the floor, we can use Newton's second law:

F = ma

where F is the force, m is the mass of the player, and a is the acceleration.

By substituting the given values, we find F ≈ 6.81 N. Therefore, the correct option is c) 6.81 m/s.